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PSLE

PSLE Foundation Mathematics

<div class="graph-block" data-graph-config="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" data-width="360" data-height="280" style="width:360px;height:280px;max-width:100%;display:block;margin:0 auto;"></div>
<p>\(X\), \(Y\) and \(Z\) are three points on horizontal ground. \(\angle ZXY=35^\circ\), \(\angle XYZ=90^\circ\) and \(XZ=20\text{ cm}\).</p>
<ol class="abc-list"><li data-list-item-id="sec4-em-2025-aiss-p1-q12-b3-i1" data-list-item-path="sec4-em-2025-aiss-p1-q12-b3-i1"><div class="list-item-paragraph">Calculate the distance \(YZ\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="sec4-em-2025-aiss-p1-q12-b3-i2" data-list-item-path="sec4-em-2025-aiss-p1-q12-b3-i2"><div class="list-item-paragraph">Calculate the area of triangle \(XYZ\).<span class="inline-marks">[2]</span></div></li></ol>

\(X\), \(Y\) and \(Z\) are three points on horizontal ground. \(\angle ZXY=35^\circ\), \(\angle XYZ=90^\circ\) and \(XZ=20\text{ cm}\).

Basics of Trigonometric Ratios

Area of Triangle

2025 AISS PRELIMS P1 Q12

Perimeter and Area of Plane Figures

Trigonometry

<p>The diagram shows a coastal reserve bounded by four survey markers \(P\), \(Q\), \(R\) and \(S\) on level ground.</p>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<p>The bearing of \(P\) from \(S\) is \(342^\circ\) and \(SP=176\text{ m}\). Marker \(Q\) is \(80\text{ m}\) due west of \(P\).</p>
<p>The angle \(PQR\) is \(117^\circ\) and the bearing of \(P\) from \(R\) is \(052^\circ\).</p>
<p>Give non-exact numerical answers correct to \(3\) significant figures unless a different degree of accuracy is stated.</p>
<ol class="abc-list"><li data-list-item-id="emath-aot-q-a" data-list-item-path="emath-aot-q-a"><div class="list-item-paragraph">Find the bearing of \(R\) from \(P\).<span class="inline-marks">[1]</span></div></li><li data-list-item-id="emath-aot-q-b" data-list-item-path="emath-aot-q-b"><div class="list-item-paragraph">Find \(\angle QPR\).<span class="inline-marks">[1]</span></div></li><li data-list-item-id="emath-aot-q-c" data-list-item-path="emath-aot-q-c"><div class="list-item-paragraph">Show that the distance \(PR\) is \(168.66\text{ m}\), correct to two decimal places.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="emath-aot-q-d" data-list-item-path="emath-aot-q-d"><div class="list-item-paragraph">Find the area of triangle \(PQR\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="emath-aot-q-e" data-list-item-path="emath-aot-q-e"><div class="list-item-paragraph">Find \(\angle RPS\).<span class="inline-marks">[1]</span></div></li><li data-list-item-id="emath-aot-q-f" data-list-item-path="emath-aot-q-f"><div class="list-item-paragraph">Hence, calculate the distance \(SR\).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="emath-aot-q-g" data-list-item-path="emath-aot-q-g"><div class="list-item-paragraph">An observation tower, \(24\text{ m}\) high, stands vertically at \(P\).</div><div class="list-item-paragraph">Find the angle of elevation of the top of the tower from \(S\).<span class="inline-marks">[2]</span></div></li></ol>

The diagram shows a coastal reserve bounded by four survey markers \(P\), \(Q\), \(R\) and \(S\) on level ground.

The bearing of \(P\) from \(S\) is \(342^\circ\) and \(SP=176\text{ m}\). Marker \(Q\) is \(80\text{ m}\) due west of \(P\).

The angle \(PQR\) is \(117^\circ\) and the bearing of \(P\) from \(R\) is \(052^\circ\).

Give non-exact numerical answers correct to \(3\) significant figures unless a different degree of accuracy is stated.

Sine Rule

Angles of Elevation and Depression

Cosine Rule

Bearings

Surveying a coastal reserve

<p>Triangle \(ABC\) is a right-angled triangle with angle \(ACB = 90^\circ\). \(ADEB\), \(BCGF\) and \(ACHI\) are squares. \(CJK\) is a straight line such that&nbsp;\(CJK\) is perpendicular to \(AB\) and \(DE\).<br>Show that</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">triangles \(DAC\) and \(BAI\) are congruent, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">the area of quadrilateral \(ADKJ\) and the area of square \(ACHI\) are equal;</span></div></li><li><div class="list-item-content"><span class="list-item-text">triangles \(BEC\) and \(BAF\) are congruent, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">the area of quadrilateral \(JKEB\) and the area of square \(BCGF\) are equal, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">\(AB^2 = AC^2 + BC^2\) (the Pythagoras Theorem).</span></div></li></ol>
<div class="content-block"><figure class="question-figure alignleft" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770799522790-6382ys9wbpl.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTk1MjI3OTAtNjM4MnlzOXdicGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.jKtUT1Eq557kZVetAj2Z5Q3HNOOiVZ4rjYCXLGkHDZ8" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignleft custom-width" /></figure></div>

Triangle \(ABC\) is a right-angled triangle with angle \(ACB = 90^\circ\). \(ADEB\), \(BCGF\) and \(ACHI\) are squares. \(CJK\) is a straight line such that \(CJK\) is perpendicular to \(AB\) and \(DE\).
Show that

triangles \(DAC\) and \(BAI\) are congruent, hence

the area of quadrilateral \(ADKJ\) and the area of square \(ACHI\) are equal;

triangles \(BEC\) and \(BAF\) are congruent, hence

the area of quadrilateral \(JKEB\) and the area of square \(BCGF\) are equal, hence

\(AB^2 = AC^2 + BC^2\) (the Pythagoras Theorem).

Area Ratios of Triangles with Equal Heights

Perimeter and Area of Rectangles and Squares

Congruence

IP2U8 HOT Q4 (Euclid's Proof of the Pythagoras' Theorem)

Congruence and Similarity

<p>In triangle \(ABC\), \(\angle BAC = 45^\circ\), \(\angle ACB = 22.5^\circ\), \(BC = 12\) cm. Find the exact area of triangle \(ABC\).</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770798105175-922gq7pr9nq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTgxMDUxNzUtOTIyZ3E3cHI5bnEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.8_NCkcB67EcYn9ZPFLs0fy7E487fkmetSQBJlbtCiKI" style="max-width: min(250px, 100%); height: auto; display: block;" alt="Triangle ABC with given angles and side length BC" class="alignright custom-width" /></figure></div>

In triangle \(ABC\), \(\angle BAC = 45^\circ\), \(\angle ACB = 22.5^\circ\), \(BC = 12\) cm. Find the exact area of triangle \(ABC\).

IP2U8 HOT Q1

<p>The figure shows a triangle \(ABC\) where&nbsp;\(BC = 23\) cm, \(K\) lies on \(BC\) such that the length of \(BK\) is \(135\%\) of the length of \(KC\) and \(\angle AKC = 90^\circ\). Given that the area of \(\Delta ABK\) is \(68\text{ cm}^2\), find \(\angle BAC\).</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770787730690-57r7qj595po.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3ODc3MzA2OTAtNTdyN3FqNTk1cG8ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.lfk0cwq1hX-dCzwz3AhDItjProkzBzuW1pv3aiF2bvA" style="max-width: min(320px, 100%); height: auto; display: block;" alt="Diagram of triangle ABC with altitude AK to base BC" class="alignright custom-width" /></figure></div>

The figure shows a triangle \(ABC\) where \(BC = 23\) cm, \(K\) lies on \(BC\) such that the length of \(BK\) is \(135\%\) of the length of \(KC\) and \(\angle AKC = 90^\circ\). Given that the area of \(\Delta ABK\) is \(68\text{ cm}^2\), find \(\angle BAC\).

Applications of Trigonometric Ratios

IP2U7 Exercise Q10

<p>In triangle \(ABC\), \(AB = 4\) cm, \(AC = 3\) cm and \(BC = 6\) cm. Triangle \(A'B'C\) is obtained by rotating triangle \(ABC\) in clockwise by \(\frac{\pi}{2}\) rad with \(C\) as the centre. Find the exact area of the shaded region.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770721740037-13c3fputrpa.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjE3NDAwMzctMTNjM2ZwdXRycGEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.fZWgbfQjbTPjp8N19OzrhpWZYE6-OYCKpLaFYVlC7-E" style="max-width: min(180px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In triangle \(ABC\), \(AB = 4\) cm, \(AC = 3\) cm and \(BC = 6\) cm. Triangle \(A'B'C\) is obtained by rotating triangle \(ABC\) in clockwise by \(\frac{\pi}{2}\) rad with \(C\) as the centre. Find the exact area of the shaded region.

IP2U13 HOT Q2

<p>The figure below is made up of two intersecting circles. The points \(A\) and \(B\) lie on the circumference of the circles. The larger circle has centre \(C\), \(AC = BC = 16\) cm and \(\angle ACB = 0.85\) radians.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Calculate the area of the shaded segment.<br>The smaller circle, with centre \(D\), cuts the first circle at \(A\) and \(B\).<br>The radius of the smaller circle is \(12\) cm.</div></li><li><div class="list-item-paragraph">Calculate, in radians, \(\angle ADB\).</div></li><li><div class="list-item-paragraph">Calculate the perimeter of the sector \(ADB\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-end;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770720293430-qozfm2oam2.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjAyOTM0MzAtcW96Zm0yb2FtMi5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0NzI1LCJleHAiOjE3OTM1MDUwMjV9.QI16HqNWdQZ1JA4D2gM9lUh1M-n4Pg3CASlKpn7GbK8" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

The figure below is made up of two intersecting circles. The points \(A\) and \(B\) lie on the circumference of the circles. The larger circle has centre \(C\), \(AC = BC = 16\) cm and \(\angle ACB = 0.85\) radians.

IP2U13 HOT Q1

<p>The diagram shows two circles, with centres \(A\) and \(B\), each of radius \(5\) cm, touching each other externally at \(X\). Both of them touch another bigger circle, with centre \(O\) and radius \(13\) cm, internally at \(D\) and \(C\) respectively.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Show that&nbsp;\(\angle COD\) is \(1.35\) radians.</div></li><li><div class="list-item-paragraph">Calculate</div><ol class="i-list"><li><div class="list-item-paragraph">the perimeter and</div></li><li><div class="list-item-paragraph">the area of the shaded region.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770718883074-uu9xtodyzsa.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MTg4ODMwNzQtdXU5eHRvZHl6c2EucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ._OhY05i6DCLcWV1jGn7enNBt-u6Y2g3YpbZ6EDUw8kI" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

The diagram shows two circles, with centres \(A\) and \(B\), each of radius \(5\) cm, touching each other externally at \(X\). Both of them touch another bigger circle, with centre \(O\) and radius \(13\) cm, internally at \(D\) and \(C\) respectively.

IP2U13 13.4 PN1 Q12

<p>In the diagram below, \(O\) is the centre of the circle with radius \(4\) cm and passing through the points \(A\), \(B\), \(C\) and \(D\). \(E\) is the midpoint of \(OC\). \(AOEC\) is a straight line. \(A\) is the centre of a circle through \(B\), \(E\) and \(D\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Show that&nbsp;\(\angle AOB\) is approximately \(1.70\) rad.</div></li><li><div class="list-item-paragraph">Calculate</div><ol class="i-list"><li><div class="list-item-paragraph">the area of sector \(OBCD\),</div></li><li><div class="list-item-paragraph">the perimeter of the shaded region,</div></li><li><div class="list-item-paragraph">the area of the shaded region.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770717368151-kxff6a7eiqb.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MTczNjgxNTEta3hmZjZhN2VpcWIucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.blNfglpWXaSo7nCtb87t2q_WpaEUhdMhdiylzliUK7E" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram below, \(O\) is the centre of the circle with radius \(4\) cm and passing through the points \(A\), \(B\), \(C\) and \(D\). \(E\) is the midpoint of \(OC\). \(AOEC\) is a straight line. \(A\) is the centre of a circle through \(B\), \(E\) and \(D\).

IP2U13 13.4 PN1 Q10

<p>The figure shows two circles with centres \(A\) and \(B\), of radii \(2\) cm and \(5\) cm respectively.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>Given that&nbsp;\(\angle PAQ = \frac{\pi}{2}\) radians,</p><ol class="abc-list"><li><div class="list-item-paragraph">show that&nbsp;\(\angle PBQ\) is approximately \(0.5735\) radians,</div></li><li><div class="list-item-paragraph">calculate the area of the shaded region.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770716485147-zfeu2cx6m2.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MTY0ODUxNDctemZldTJjeDZtMi5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0NzI1LCJleHAiOjE3OTM1MDUwMjV9.Cx2PLNvx2EDwyNp6vJaP2ODrT9hPvvOhF3oILIU6fAY" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

The figure shows two circles with centres \(A\) and \(B\), of radii \(2\) cm and \(5\) cm respectively.

IP2U13 13.4 PN1 Q9

<p>The diagram shows a sector of a circle with centre \(O\) and radius \(8\) cm.</p>
<p>\(P\) is a point on \(OA\) such that&nbsp;\(4OP = 3OA\) and \(Q\) is a point on \(OB\) such that&nbsp;\(5OQ = 2OB\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>Given that&nbsp;\(\angle AOB = 0.8\) radians, calculate</p><ol class="i-list"><li><div class="list-item-paragraph">the length of \(PQ\),</div></li><li><div class="list-item-paragraph">the arc length \(AB\),</div></li><li><div class="list-item-paragraph">the area of the shaded region.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770715613188-j6lcrix7ouc.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MTU2MTMxODgtajZsY3JpeDdvdWMucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.G_gmrOftf4w--TlxaBuPrwEMGd7eErn0WNDgOqUoJjY" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

The diagram shows a sector of a circle with centre \(O\) and radius \(8\) cm.

\(P\) is a point on \(OA\) such that \(4OP = 3OA\) and \(Q\) is a point on \(OB\) such that \(5OQ = 2OB\).

IP2U13 13.4 PN1 Q8

<p>In triangle \(ABC\), \(a\), \(b\) and \(c\) are the opposite sides of the interior angles \(A\), \(B\) and \(C\), respectively.</p>
<p>Let&nbsp;\(s = \frac{a + b + c}{2}\), show that the area of triangle is given by \(\sqrt{s(s - a)(s - b)(s - c)}\).</p>

In triangle \(ABC\), \(a\), \(b\) and \(c\) are the opposite sides of the interior angles \(A\), \(B\) and \(C\), respectively.

Let \(s = \frac{a + b + c}{2}\), show that the area of triangle is given by \(\sqrt{s(s - a)(s - b)(s - c)}\).

Heron's Formula

<p>In triangle \(ABC\), \(a\), \(b\) and \(c\) are the opposite sides of the interior angles \(A\), \(B\) and \(C\), respectively.</p>
<p>It is given that&nbsp;\(c = 2\) and \(b = \sqrt{2}a\), show that the maximum area of triangle \(ABC\) is \(2\sqrt{2}\).</p>
<p>Hence find the exact value of \(a\) when the area of triangle is maximised.</p>

It is given that \(c = 2\) and \(b = \sqrt{2}a\), show that the maximum area of triangle \(ABC\) is \(2\sqrt{2}\).

Hence find the exact value of \(a\) when the area of triangle is maximised.

IP2 U12 HOT Q3

<p>The diagram shows a sheet of cardboard \(ABCD\).</p>
<p>\(AD = 10\) cm, \(\angle ABC = 46^\circ\), \(\angle BAC = 75^\circ\), \(\angle ACD = 33^\circ\) and \(\angle ADC = 90^\circ\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Calculate</div><ol class="i-list"><li><div class="list-item-paragraph">\(AC\),</div></li><li><div class="list-item-paragraph">\(BC\) and,</div></li><li><div class="list-item-paragraph">the area of the triangle \(ABC\).</div></li></ol></li><li><div class="list-item-paragraph">Triangle \(ADC\) is folded along \(AC\) to form triangle \(AEC\).<br>Calculate the length of</div><ol class="i-list"><li><div class="list-item-paragraph">\(EC\) and,</div></li><li><div class="list-item-paragraph">\(BE\).</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770533029291-u2ypx5enu1.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA1MzMwMjkyOTEtdTJ5cHg1ZW51MS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0NzI1LCJleHAiOjE3OTM1MDUwMjV9.GKRmZ2_A-1asr7bQayJOA5oNjculDxWWZ8ieciaXkrA" style="max-width: min(280px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

The diagram shows a sheet of cardboard \(ABCD\).

\(AD = 10\) cm, \(\angle ABC = 46^\circ\), \(\angle BAC = 75^\circ\), \(\angle ACD = 33^\circ\) and \(\angle ADC = 90^\circ\).

IP2U12 Exercise Trigonometry II (also for 3EM U7 Worksheet) Q11

<p>In the diagram, \(BCD\) is a straight line, \(BC = 6\) cm, \(AC = 7.9\) cm, \(AD = 6.4\) cm and \(\angle ACB = 130^\circ\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>Calculate</p><ol class="i-list"><li><div class="list-item-paragraph">\(AB\),</div></li><li><div class="list-item-paragraph">the area of triangle \(ABC\),</div></li><li><div class="list-item-paragraph">the shortest distance from \(B\) to \(AC\) produced,</div></li><li><div class="list-item-paragraph">\(\angle ADC\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: center;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770532306077-rvbo3a2qk29.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA1MzIzMDYwNzctcnZibzNhMnFrMjkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.NiKEmFXBJD_CmU85ktpHZ6uBFAquoJOKztmW3tKKczY" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, \(BCD\) is a straight line, \(BC = 6\) cm, \(AC = 7.9\) cm, \(AD = 6.4\) cm and \(\angle ACB = 130^\circ\).

IP2U12 Exercise Trigonometry II (also for 3EM U7 Worksheet) Q10

<p>In the diagram, \(AB = 4\) cm, \(BC = 7.5\) cm, \(AD = 5\) cm, \(BD = 6.2\) cm and \(\angle BDC = 55^\circ\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>Calculate</p><ol class="i-list"><li><div class="list-item-paragraph">\(\angle ADB\),</div></li><li><div class="list-item-paragraph">\(\angle BCD\),</div></li><li><div class="list-item-paragraph">the area of \(ABCD\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770531648311-ftgm9c3c6f5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA1MzE2NDgzMTEtZnRnbTljM2M2ZjUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.-DRZEfAC8wIyfjAmaJcNK_2cPc-Df5Zs9LrECGlhUH0" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, \(AB = 4\) cm, \(BC = 7.5\) cm, \(AD = 5\) cm, \(BD = 6.2\) cm and \(\angle BDC = 55^\circ\).

<p>The diagram shows a quadrilateral \(ABCD\).</p>
<p>The length of \(AB\) is \(5\) m, the length \(BC\) is \(8\) m and the length of \(AD\) is \(4\) m.</p>
<p>Angle \(ABC\) is \(60^\circ\) and angle \(ADC\) is \(100^\circ\).</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Calculate the length of \(AC\).</span></div></li><li><div class="list-item-content"><span class="list-item-text">Calculate angle \(ACD\).</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the area of the quadrilateral \(ABCD\), correct to the nearest whole number.</span></div></li></ol>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770530950473-g12h1z2uknv.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA1MzA5NTA0NzMtZzEyaDF6MnVrbnYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.JqP5IT3wxDx2ZS5Wx4oC6IesYf89MAE7cjiZS_-iFJg" style="max-width: min(220px, 100%); height: auto; display: block;" alt="Diagram of quadrilateral ABCD" class="aligncenter custom-width" /></figure></div>

The diagram shows a quadrilateral \(ABCD\).

The length of \(AB\) is \(5\) m, the length \(BC\) is \(8\) m and the length of \(AD\) is \(4\) m.

Angle \(ABC\) is \(60^\circ\) and angle \(ADC\) is \(100^\circ\).

Find the area of the quadrilateral \(ABCD\), correct to the nearest whole number.

IP2U12 Exercise Trigonometry II (also for 3EM U7 Worksheet) Q8

<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>In the diagram, \(AB = 6\) cm, \(BC = 4\) cm and \(AC = 5\) cm.<br>Without finding any angle, calculate</p><ol class="i-list"><li><div class="list-item-paragraph">the value of \(\cos C\).</div></li><li><div class="list-item-paragraph">the value of ratio \(\frac{\sin A}{\sin C}\).</div></li><li><div class="list-item-paragraph">find \(\sin B\) if the area of \(\Delta ABC = 9.9\text{ cm}^2\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770530337592-qftt32i4f4f.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA1MzAzMzc1OTItcWZ0dDMyaTRmNGYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDcyNSwiZXhwIjoxNzkzNTA1MDI1fQ.9I_4BC8Wg5SN-4I3--c4dGsUv72bvqiKpdaxc3riHf4" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

IP2U12 Exercise Trigonometry II (also for 3EM U7 Worksheet) Q7

<p>In the diagram, \(BDCE\) is a straight line, \(BD = 4\) cm, \(AC = 10\) cm and \(AB = AD\).</p>
<p>Given that the area of \(\Delta ABD = 16\) cm\(^2\), find the value of \(\sin \angle ACD\).</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770526492553-v1727jwj35.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA1MjY0OTI1NTMtdjE3Mjdqd2ozNS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0NzI1LCJleHAiOjE3OTM1MDUwMjV9.SPbag9cXzt7QgU5F82fZAGW9LTZCRmHCllIlbEPl0Q8" style="max-width: min(300px, 100%); height: auto; display: block;" alt="Geometry diagram showing triangle ABC with point D on BC." class="aligncenter custom-width" /></figure></div>

In the diagram, \(BDCE\) is a straight line, \(BD = 4\) cm, \(AC = 10\) cm and \(AB = AD\).

Given that the area of \(\Delta ABD = 16\) cm\(^2\), find the value of \(\sin \angle ACD\).

IP2U12 Exercise Trigonometry II (also for 3EM U7 Worksheet) Q4

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<p>In the diagram, \(AB = 16\) cm, \(CD = 18\) cm, \(\angle ABC = 90^\circ\), \(\cos \angle BAC = \frac{4}{5}\) and \(BCDE\) is a straight line.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find \(AC\),</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find \(\tan \angle ACB\),</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find \(AD\),</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find \(\angle ADE\),</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the shortest distance from \(C\) to \(AD\).</span><span class="inline-marks">[3]</span></div></li></ol>

In the diagram, \(AB = 16\) cm, \(CD = 18\) cm, \(\angle ABC = 90^\circ\), \(\cos \angle BAC = \frac{4}{5}\) and \(BCDE\) is a straight line.

Find the shortest distance from \(C\) to \(AD\).

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